Question

A cuboid shaped slab of iron whose dimensions are $55cm\times 40cm\times 15cm$ is melted and reacast into a pipe. The outer diameter and thickness of the pipe are $8cm$ and $1cm$ respectively. Find the length of the pipe. (Take $\pi =\frac{22}{7}$)

Answer 1

Let $h_1$ be the length of the pipe.

Let $R$ and $r$ be the outer and inner radii of the pipe respectively.

Iron slab:

Length $=55cm$

Breadth $=40cm$

Height $=15cm$

Volume $=lbh=55\times 40\times 15c{m}^3$

Iron pipe:

Outer diameter, $2R=8cm$

$\therefore$ Outer radius, $R=4cm$

Thichness, $w=1cm$

$\therefore$ Inner radius $r=R-w=(4-1)cm=3cm$

Now,

The volume of the iron pipe $=$ Volume of iron slab

$\Rightarrow \pi h_1(R^2-r^2)=lbh$

$\Rightarrow \frac{22}{7}\times h_1(4^2-3^2)=55\times 40\times 15$

$\Rightarrow \frac{22}{7}\times h_1\times 7=55\times 40\times 15$

$\Rightarrow 22\times h_1=55\times 40\times 15$

$\Rightarrow h_1=\frac{55\times 40\times 15}{22}$

$\Rightarrow h_1=1500$

Thus, the length of the pipe, $h_1=1500cm=15m$